ThermochemistryHard

Question

A volume of 100 ml of 0.5 N-H2SO4 solution is neutralized with 200 ml of 0.2 M-NH4OH in a constant pressure calorimeter which resulted 1.4o C rise in temperature. The heat capacity of the calorimeter system is 1.5kJ/oC. Some useful thermochemical equations are as follows:

HCl + NaOH → NaCl + H2O + 57 kJ

CH3COOH + NH4OH → CH3COONH4 + H2O + 48.1 kJ

Which of the following statements are correct?

Options

A.Enthalpy of neutralization of HCl vs. NH4OH is −52.5 kJ/mol.
B.Enthalpy of dissociation (ionization) of NH4OH is 4.5 kJ/mol.
C.Enthalpy of dissociation (ionization) of CH3COOH is 4.6 kJ/mol.
D.ΔH for 2H2O(l) → 2H+ (aq) + 2OH− (aq) is 114 kJ.

Solution

Heat released in reaction = Heat gained by calorimeter system $= 1.5 \times 1.4 = 2.1\text{ kJ}$

$n_{eq}\left( H_{2}SO_{4} \right) = \frac{100 \times 0.5}{1000} = 0.05 $$${n_{eq}\left( NH_{4}OH \right) = \frac{200 \times 0.2}{1000} = 0.04 \left( \text{Limiting reagent} \right) }{\Delta_{\text{neut}}H_{NH_{4}OH}\left( \text{By strong acid} \right) = - \frac{2.1}{0.04} = - 52.5\text{ kJ/eq} }{\Delta_{\text{diss}}H_{NH_{4}OH} = ( - 52.5) - ( - 57) = 4.5\text{ kJ/mol} }{\Delta_{\text{diss}}H_{CH_{3}COOH} = (57 - 48.1) - 4.5 = 4.4\text{ kJ/mol}}$$

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